2008 Physics. Standard Grade Credit. Finalised Marking Instructions

Transcription

1 2008 Physics Standard Grade Credit Finalised Marking Instructions Scottish Qualifications Authority 2008 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from the Assessment Materials Team, Dalkeith. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre's responsibility to obtain the necessary copyright clearance. SQA's Assessment Materials Team at Dalkeith may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 Physics Marking Issues The current in a resistor is 1 5 amperes when the potential difference across it is 7 5 volts. Calculate the resistance of the resistor. Answers Mark + Comment Issue 1. V=IR (½) Ideal answer 7 5=1 5R (½) R=5 0 Ω (1) Ω (2) Correct answer GMI (1½) Unit missing GMI 2 (a) Ω (0) No evidence/wrong answer GMI 1 5. Ω (0) No final answer GMI 1 V R = = = 4 0 Ω (1½) Arithmetic error GMI 7 I V R = = 4 0 Ω I (½) Formula only GMI 4 and 1 V 8. R = = Ω (½) Formula only GMI 4 and 1 I 9. V 7 5 R = = = I 1 5 Ω (1) Formula + subs/no final answer GMI 4 and V 7 5 R = = = 4 0 I 1 5 (1) Formula + substitution GMI 2 (a) and V 1 5 R = = = 5 0 Ω I 7 5 (½) Formula but wrong substitution GMI V 75 R = = = 5 0 Ω I 1 5 (½) Formula but wrong substitution GMI I 7 5 R = = = 5 0 Ω V 1 5 (0) Wrong formula GMI V = IR 7 5 = 1 5 R R = 0 2 Ω (1½) Arithmetic error GMI V = IR I 1 5 R = = = 0 2 Ω V 7 5 (½) Formula only GMI 20 Page 2

3 DATA SHEET Speed of light in materials Speed of sound in materials Material Speed in m/s Material Speed in m/s Air Aluminium 5200 Carbon dioxide Air 340 Diamond Bone 4100 Glass Carbon dioxide 270 Glycerol Glycerol 1900 Water Muscle 1600 Steel 5200 Tissue 1500 Water 1500 Gravitational field strengths Specific heat capacity of materials Gravitational field strength on the surface in N/kg Material Specific heat capacity in J/kg C Earth 10 Alcohol 2350 Jupiter 26 Aluminium 902 Mars 4 Copper 386 Mercury 4 Glass 500 Moon 1 6 Glycerol 2400 Neptune 12 Ice 2100 Saturn 11 Lead 128 Sun 270 Silica 1033 Venus 9 Water 4180 Specific latent heat of fusion of materials Melting and boiling points of materials Material Specific latent heat of fusion in J/kg Material Melting point in C Boiling point in C Alcohol Alcohol Aluminium Aluminium Carbon dioxide Copper Copper Glycerol Glycerol Lead Lead Turpentine Water Specific latent heat of vaporisation of materials SI Prefixes and Multiplication Factors Material Specific latent heat Prefix Symbol Factor of vaporisation in J/kg Alcohol giga G = 10 9 Carbon dioxide mega M = 10 6 Glycerol kilo k 1000 = 10 3 Turpentine milli m = 10-3 Water micro µ = 10-6 nano n = 10-9 Page 3

4 K&U PS Marks 1. A high definition television picture has 1080 lines and there are 25 pictures produced each second. (a) (i) Calculate how long it takes to produce one picture on the screen. Space for working and answer (time for one picture = 1 s) 25 = 0 04 s (1) deduct (½) if no unit 1 (ii) Explain why a continuous moving picture is seen on the television screen and not 25 individual pictures each second. new image is different from previous image (1) the brain retains the image (1) (while the new image is being displayed) 2 (b) The television picture is in colour. (i) Which two colours are used to produce magenta on the screen? red and blue (1) for both 1 (ii) Due to a fault, the colour yellow appears as orange on the screen. Which colour should be reduced in brightness to correct this problem? red (1) 1 Page 4

5 NOTES (i) If wrong final answer but correct substitution deduct (½) 1 Accept: s 25 (ii) Look for answer connecting following two statements the image is retained short time (interval) between each image OR each image is different (or phrase that suggests this) statements could be in different order Question No 1

6 K&U PS Marks 2. A television company is making a programme in China. Britain receives television pictures live from China. The television signals are transmitted using microwaves. The microwave signals travel from China via a satellite, which is in a geostationary orbit. (a) State what is meant by a geostationary orbit. period is 24 hours OR always above the same point above the Earth OR period same as Earth 1 (b) The diagram shows the position of the transmitter and receiver. Complete the diagram to show the path of the microwave signals from China to Britain. (1) for showing satellite in position (label not required) (1) for correct path direction lines may be (0) marks for straight line between slightly curved China and Britain Britain Receiver China Transmitter (c) The frequency of the microwave signals being used for transmission is 8 GHz. 2 (i) What is the speed of the microwaves? m/s (1) or (0) unit required 1 (ii) Calculate the wavelength of these microwaves. Space for working and answer λ = v f (½) = (½) = m (1) 2 Page 5

7 NOTES (a) Accept: same rate of rotation as Earth km orbit Do not accept: same speed as Earth same point (in space) stationary (b) (1) for a V shape (indicates re-transmission) (1) for direction arrow or arrows (C to B) not necessary to show or label satellite 0 marks for straight line between China and Britain (even with arrow) BUT may see a straight line with a box or circle between China/Britain which represents satellite if direction arrow included (2) marks ALSO could be 2 satellites shown with signals connecting in space deduct (½) if no/wrong conversion sig fig range: Question No 2

8 K&U PS Marks 3. In a sprint race at a school sports day, the runners start when they hear the sound of the starting pistol. An electronic timer is also started when the pistol is fired into the air. electronic timer lane 1 lane 2 lane 3 lane 4 lane 5 lane m Not to scale 10 m The runner in lane 1 is 3 2 m from the starting pistol. The runner in lane 6 is 10 m from the starting pistol. (a) The runner in lane 1 hears the starting pistol first. Calculate how much later the runner in lane 6 hears this sound after the runner in lane 1. Space for working and answer t = d v (½) = (½) = 0 02 s (1) (1) for data selection of 340 (m/s) 3 if any other value for v from speed of sound in materials table (2) max still possible any other value of v (½) max only Page 6

9 NOTES (a) alternative 2 stages (½) formula (½) both workings (1) final answer d = 3 2 m d = 10 m so t = d d t = t = v v = (accept 0 02 or s) = = if more sig figs then deduct (½) = (s) = (s) Question No 3 (a)

10 K&U PS Marks (b) A sensor detects each runner crossing the finishing line to record their time. The table gives information about the race. Place Lane Time (s) 1st nd rd Using your answer to part (a), explain why the runner in lane 6 should have been awarded first place. Space for working and answer (1) for showing a correction/appreciation of times for runners 1 and 6 (1) for comparison to show that 6 is winner 2 (c) One runner of mass 60 kg has a speed of 9 m/s when crossing the finishing line. Calculate the kinetic energy of the runner at this point. Space for working and answer E K = 1 2 mv 2 (½) = (½) = 2430 J (1) 2 Page 7

11 NOTES (b) Candidates with t less than can still get (1) out of (2) for showing correction/appreciation of times if no square at 9 2 substitution error then (½) max Question No 3 (b) and (c)

12 K&U PS Marks 4. A student has four resistors labelled A, B, C and D. The student sets up Circuit 1 to identify the value of each resistor. 6 0 V A resistor V Circuit 1 Each resistor is placed in the circuit in turn and the following results are obtained. Resistor Voltage across resistor (V) Current (A) A B C D (a) (i) Show, by calculation, which of the resistors has a value of 120 Ω. Space for working and answer V V I = (½) OR V = IR (½) OR R = (½) R I 6 6 = (½) = (½) = (½) = 0 05(A) (1) = 6(V) (1) = 120(Ω) (1) ie it is resistor C (1) must show working to obtain final mark no sig. fig. issue in calculation 3 Page 8

13 NOTES may see sequential calculations until 120 Ω is reached ignore other calculations even if in error Question No 4 (a) (i)

14 K&U PS Marks (ii) The student then sets up Circuit 2 to measure the resistance of each resistor. 120 Ω OFF V Ω resistor A Circuit 2 State one advantage of using Circuit 2 to measure the resistance compared to using Circuit 1. direct reading OR no need for calculation OR easier (to obtain answer) or quicker 1 (b) The resistances of the other three resistors are 180 Ω, 220 Ω and 360 Ω. The student collects all four resistors in series. Calculate the total resistance. Space for working and answer R = R 1 + R 2 + R3 + R 4 (½) = (½) = 880 Ω (1) 2 Page 9

15 NOTES Do not accept: more accurate Accept: R = R 1 + R 2 as formula if only 3 resistors added then (½) max for (implied) formula Question No 4 (a) (ii) and (b)

16 K&U PS Marks 5. The diagram shows three household circuits connected to a consumer unit. lighting circuit cooker socket 30 A 5 A 30 A fuse fuse fuse consumer unit ring circuit socket socket (a) (i) State one advantage of a ring circuit. cheaper/less current in each branch/thinner wire 1 (ii) State the value of mains voltage. 230 V (1) or (0) unit required 1 (b) Each of the lamps in the lighting circuit has a power rating of 100 W. One of the lamps is switched on. (i) Calculate the current in the lamp. Space for working and answer I = P V (½) = (½) = A (1) sig. figs.: Page 10

17 NOTES (i) Accept: convenient for adding extra sockets 2 paths for current less voltage drop less heat Do not accept: less wire less current per ring any comparison with series circuit safer Question No 5 (a) and (b) (i)

18 K&U PS Marks (b) (ii) Explain why a house with twenty 100 W lamps requires two separate lighting circuits. the current/power is too high (1) so fuse could break if only one 2 circuit (1) OR the fuse value is not enough (1) Page 11

19 NOTES 1 statement about power or current 1 conclusion about fuse size being insufficient eg too much current (implied >5 A (1)) so fuse breaks (1) OR there is too much current/power (1) so too big for one fuse (1) Do not accept: too much power for lights there is not a big enough fuse for all 20 Question No 5 (b) (ii)

20 K&U PS Marks 6. A short-sighted person has difficulty seeing the picture on a cinema screen. Figure 1 shows rays of light from the screen entering an eye of the person until the rays reach the retina. light rays from screen retina Figure 1 (a) (i) In the dotted box in Figure 2, draw the shape of the lens that would correct this eye defect. 1 light rays from screen concave lens only (1) (1) for showing divergence (1) for showing convergence on retina retina Figure 2 (ii) In figure 2, complete the path of the rays of light from this lens until they reach the retina. (a) (i) independent from (ii) 2 Page 12

21 NOTES rays may start converging at cornea ok ignore rays drawn inside dotted box ok Question No 6 (a)

22 K&U PS Marks (b) Doctors can use an endoscope to examine internal organs of a patient. The endoscope has two separate bundles of optical fibres that are flexible. light source fibre bundle X eye tip of endoscope fibre bundle Y A section of optical fibre used in the endoscope is shown below. (i) Complete the diagram to show how light is transmitted along the optical fibre. (1) for showing total internal reflection 2 (1) for quality (no more than 6 reflections) (ii) Explain the purpose of each bundle of optical fibres in the endoscope. Fibre bundle X Fibre bundle Y transmits the light inside the patient transmits the (reflected) light/image back to the doctor s eye 2 (iii) The tip of the endoscope that is inside the patient is designed to be very flexible. Suggest one reason for this. to allow the doctor to see/steer/manoeuvre to different parts of inside the body/the organ 1 Page 13

23 NOTES (b) (i) (1) mark for quality lost if: not passably straight lines not passably equal <i = <r note: arrows, normals NOT required (ii) look for the following principles: X Y light into are keywords light/image out are keywords do not accept: to let the doctor see inside the patient (iii) Do not accept: to stop getting jammed if rigid would cause damage Question No 6 (b)

24 K&U PS Marks 7. A hospital technician is working with a radioactive source. The graph shows the activity of the source over a period of time activity in kbq time in hours (a) (i) State what is meant by the term half-life. (ii) time for the activity (or number of nuclei) (of a radioactive 1 source) to reduce to half the original number/activity/its value Use information from the graph to calculate the half-life of the radioactive source. Space for working and answer Activity kbq 6 hours (1) deduct (½) if no/wrong unit 1 Page 14

25 NOTES Do not accept: count rate radio activity radiation strength power something Accept abbreviations: hrs or h Question No 7 (a) (i) and (ii)

26 K&U PS Marks (iii) The initial activity of the source is 160 kbq. Calculate the activity, in kbq, of the radioactive source after four halflives. Space for working and answer (½) for halving So activity is 10 (kbq) (½) (unit not required) if wrong unit given deduct (½) 1 (b) As a safety precaution the technician wears a film badge when working with radioactive sources. The film badge contains photographic film. Light cannot enter the badge. uncovered window lead 1 mm thickness aluminium 3 mm thickness Describe how the film badge indicates the type and amount of radiation received. windows allow different radiations to pass through (1) film becomes fogged/blackened/darkened (1) 2 Page 15

27 NOTES look for: answer relating to type of radiation answer relating to amount of radiation Question No 7 (a) (iii) and (b)

28 K&U PS Marks 8. A torch contains five identical LEDs connected to a 3 0 V battery as shown. S 3 0 V (a) State the purpose of the resistor connected in series with each LED. to limit current (flowing through LED) OR prevent damage to LED 1 (b) When lit, each LED operates at a voltage of 1 8 V and a current of 30 ma. (i) Calculate the value of the resistor in series with each LED. Space for working and answer V R = = 1 2 (V) (1) R = V I (½) = (½) = 40 Ω (1) (½) max if 3 V or 1 8 V used 3 (ii) Calculate the total current from the supply when all five LEDs are lit. Space for working and answer I = 5 30 = 150 ma (1) deduct (½) if no/wrong unit 1 Page 16

29 NOTES (a) Accept: protect the LED reduce voltage across LED Do not accept: to reduce voltage alone the voltage through to stop LED blowing to reduce charge/power to LED to prevent LED overheating (b) (i) if error in subtraction for V can still get (2) marks (ii) if there is a clear arith error then deduct (½) eg I = 5 30 = 160 ma (½) if wrong conversion then (½) for correct arith Question No 8 (a), (b) (i) and (ii)

30 K&U PS Marks (iii) Calculate the power supplied by the battery when all five LEDs are lit. Space for working and answer P = IV (½) = (½) = 0 45 W (1) unit penalty if wrong conversion to W deduct (½) 2 (c) State one advantage of using five LEDs rather than a single filament lamp in the torch. uses less energy/power OR more efficient OR if one fails others stay on 1 Page 17

31 NOTES Accept: less current lasts longer Do not accept: brighter less fragile Question No 8 (b) (iii) and (c)

32 K&U PS Marks 9. An electronic device produces a changing light pattern when it detects music, but only when it is in the dark. The device contains the logic circuit shown. music sensor X Z Y light sensor The music sensor produces logic 1 when the music is on and logic 0 when the music is off. The light sensor produces logic 1 when it detects light and logic 0 when it is dark. (a) (i) Suggest a suitable input device for the light sensor. LDR OR solar cell OR photodiode accept correct symbol 1 (ii) Complete the truth table for the logic levels at points X, Y and Z in the circuit. (1) for each correct column Music Light level X Y Z off dark off light on dark on light Page 18

33 NOTES Question No 9 (a)

34 K&U PS Marks (b) The device detects music from a CD player. The CD player contains an amplifier that produces an output voltage of 5 6 V when connected to a loudspeaker of resistance 3 2 Ω. (i) Calculate the output power of the amplifier. Space for working and answer (½) for both P = V 2 (½) OR I = R V R P = IV = (½) = = = 9 8 W (1) = 1 75 (A) (½) = 9 8 W (1) 2 (ii) The input power to the amplifier is 4 9 mw. Calculate the power gain of the amplifier. Space for working and answer P gain = = P P out in (½) (½) = 2000 (1) deduct (½) if unit given 2 (iii) One particular signal from the CD to the amplifier has a frequency of 170 Hz. What is the frequency of the output signal from the amplifier? 170 (Hz) accept the same 1 unit not required not: 170 MHz Page 19

35 NOTES (ii) if no conversion from 4 9 mw then unit penalty Question No 9 (b)

36 K&U PS Marks 10. A railway train travels uphill between two stations. Information about the train and its journey is given below. average speed of train time for journey power of train mass of train plus passengers 5 m/s 150 s 120 kw kg (a) Calculate the energy used by the train during the journey. Space for working and answer E = P t (½) = (½) = J (1) unit penalty if no conversion deduct (½) 2 Page 20

37 NOTES Question No 10 (a)

38 K&U PS Marks (b) Calculate the height gained by the train during the journey. Space for working and answer h = E p mg (½) = (½) = 90 m (1) 2 (c) Suggest why the actual height gained by the train is less than the value calculated in part (b). energy is lost as heat energy OR frictional heat losses OR energy is lost because of friction 1 Page 21

39 NOTES Do not accept: air resistance alone energy losses alone friction Question No 10 (b) and (c)

40 K&U PS Marks 11. A windsurfer takes part in a race. The windsurfer takes 120 seconds to complete the race. The total mass of the windsurfer and the board is 90 kg. The graph shows how the speed of the windsurfer and board changes with the time during part of the race speed in m/s time in s Page 22

41 NOTES

42 K&U PS Marks (a) (i) Calculate the acceleration of the windsurfer and board during the first 4 s of the race. Space for working and answer a = = v - u t (½) (½) = 1 25 m/s 2 (1) 2 (ii) Calculate the unbalanced force causing this acceleration. Space for working and answer F = ma (½) = (½) = N (1) sig fig range: 110 N N 113 N 2 (b) Calculate the total distance travelled by the windsurfer during the 12 s time interval shown on the graph. Space for working and answer distance travelled = area under graph (½) (or implied) 1 1 = ( 8 5) (½) for correct substitution 2 2 = ( ) = 63 5 m (1) 2 (c) What can be said about the horizontal forces acting on the windsurfer between 4 s and 6 s? forces are balanced (not equal ) OR forces are equal and opposite 1 Page 23

43 NOTES Accept units: m/s/s m s -2 Do not accept units: mpsps Question No 11 (a) (b) and (c)

44 K&U PS Marks 12. An underwater generator is designed to produce electricity from water currents in the sea. The output power of the generator depends on the speed of the water current as shown in Graph 1. 3 Graph 1 output power 2 in MW speed of water current in m/s The speed of the water current is recorded at different times of the day shown in Graph 2. 8 Graph 2 6 speed of water current 4 in m/s :00 03:00 06:00 09:00 12:00 15:00 18:00 21:00 00:00 time of day Page 24

45 NOTES

46 K&U PS Marks (a) (i) State the output power of the generator at 09: MW deduct (½) if no/wrong unit 1 (ii) State one disadvantage of using this type of generator. (power) output not consistent OR expensive OR dangerous to build/ maintain OR few suitable locations 1 (b) The voltage produced by the generator is stepped-up by a transformer. At one point in the day the electrical current in the primary coils of the transformer is 900 A and the voltage is 2000 V. The transformer is 96% efficient. (i) Calculate the output power of the transformer at this time. Space for working and answer P = IV (½) P out = = = Efficiency P in (½) (½) = 1,800,000 (W) (½) = 1,728,000 W (1) 3 (ii) State one reason why a transformer is not 100% efficient. heat losses in coils OR magnetic losses in core OR eddy currents in core 1 Page 25

47 NOTES (a) (ii) Accept: current varies variable output water speed changeable hazard to shipping/wildlife input power variable no output at times Do not accept: waves vary (ie not calm sea ) (b) (i) must calculate 96% of power ie not 96% of I or V (b) (ii) look for answer giving loss and location eg: transformer hum heat loss in core/windings energy loss in wires magnetisation of core Accept: hysteresis Do not accept: friction and heat Question No 12 (a) and (b)

48 K&U PS Marks (c) Three different types of electrical generator X, Y and Z are tested in a special tank with a current of water as shown to find out the efficiency of each generator. electrical water rotor blades generator X X speed of water current = 1 8 m/s water rotor blades electrical generator Y speed of water current = 1 9 m/s Y water rotor blades electrical generator Z speed of water current = 1 8 m/s Z Give two reasons why this is not a fair test. different water speeds OR different sizes of rotor blades OR different number of rotor blades (½) each correct 1 if one reason is different types of rotor blades (½) then the other reason must be different water speeds (½) Page 26

49 NOTES Question No 12 (c)

50 K&U PS Marks 13. In the reactor of a nuclear power station, neutrons split uranium nuclei to produce heat in what is known as a chain reaction. (a) Explain what is meant by the term chain reaction. in each reaction more neutrons are released (1) these cause further reactions (1) 2 (b) In the nuclear power station, 1 kg of uranium fuel produces MJ of heat. In a coal-fired power station 1 kg of coal produces 28 MJ of heat. Calculate how many kilograms of coal are required to produce the same amount of heat as 1 kg of uranium. Space for working and answer No of kg = = (kg) (1) unit not required if wrong unit given deduct (½) 1 (c) A power station uses an a.c. generator to convert kinetic energy from a turbine into electrical energy. A diagram of an a.c. generator is shown. stator coils rotating electromagnetic coils (i) electrical output Explain how the a.c. generator works. (electromagnetic) coils produce moving/changing magnetic field (1) voltage/current is induced in the (stator) coils (1) 2 (ii) State two changes that can be made to the generator to increase the output power. Change 1: increase rate/speed of rotation (1) Change 2: increase number of coils in (stator) OR (field) coils (1) 2 Page 27

51 NOTES (a) look for: neutrons causing fission further reactions resulting accept labelled diagram (b) if there is a clear arith error then deduct (½) (c) (i) look for: indication of changing magnetic field indication of voltage induced beware of simply a repeat of information given in question/diagram (ii) accept: increase current in rotating electromagnetic coils increase magnetic field use stronger electromagnet if use bigger, stronger magnet use ± rule do not accept: bigger/larger coils bigger magnets apply ± rule if more than 2 answers Question No 13 (a) (b) and (c)

52 K&U PS 14. A team of astronomers observes a star 200 light-years away. Marks (a) State what is meant by the term light-year. distance travelled by light in one year 1 (b) Images of the star are taken with three different types of telescope as shown. Telescope A visible light Telescope B infrared Telescope C X-ray (i) Explain why different types of telescope are used to detect signals from space. different detectors (1) are required for different radiations/ frequencies/wavelengths (1) 2 (ii) Place the telescopes in order of the increasing wavelength of the radiation which they detect. C A B (1) or (0) 1 accept: X-ray Visible Infrared (iii) State a detector that could be used in telescope C. G-M tube OR photographic film 1 (c) Telescope A is a refracting telescope with an objective lens of focal length 400 mm and diameter 80 mm. (i) Calculate the power of the objective lens. Space for working and answer 1 P = (½) f = (½) = 2 5 D (1) 2 Page 28

53 NOTES (i) look for: an indication of different radiations so different detectors are needed eg some radiations are invisible so we need something other than the eye to detect it Accept: Charge Coupled Device if no conversion to metres (P = D then unit error) accept D or d ignore ± signs Question No 14 (a) (b) and (c) (i)

54 K&U PS Marks (ii) One of the astronomers suggests replacing the objective lens in this telescope with one of larger diameter. State an advantage of doing this. fainter objects can be observed OR telescope gathers more light 1 Page 29

55 NOTES Accept: more detail clearer brighter image Do not accept: bigger picture the more you can see more focussed sharper Question No 14 (c) (ii)

56 K&U PS Marks 15. (a) A spacecraft is used to transport astronauts and equipment to a space station. On its return from space the spacecraft must re-enter the Earth s atmosphere. The spacecraft has a heat shield made from special silica tiles to prevent the inside from becoming too hot. (i) Why does the spacecraft increase in temperature when it re-enters the atmosphere? friction (between craft and atmosphere causes heat production) 1 (ii) The mass of the heat shield is kg and the gain in heat energy of the silica tiles is 4 7 GJ. Calculate the increase in temperature of the silica tiles. Space for working and answer ΔT = E H (1) for selection of 1033 (J/Kg C) mc if wrong value for c selected from = Specific heat capacity of 3 materials table then can continue with this value (2 max) = 1300 C any other value of c then (½) max 3 (iii) Explain why the actual temperature rise of the silica tiles is less than the value calculated in (a) (ii). some heat (generated) is lost to surroundings OR some heat energy reached the rest of the spacecraft 1 (b) When a piece of equipment was loaded on to the spacecraft on Earth, two people were required to lift it. One person was able to lift the same piece of equipment in the Space Station. Explain why one person was able to lift the equipment in the Space Station. weighs less (in space) 1 [END OF MARKING INSTRUCTIONS] Page 30

57 NOTES (i) Accept: because of the friction air resistance (b) Accept: the force of gravity is less Do not accept: no gravity less gravity lighter Question No 15